Question -
Answer -
Given:
The verticesof ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).
Now let us find theslopes of ∆ABC.
Slope of AB = [(2 – 4)/ (-3-1)]
= ½
Slope of BC = [(-3 –2) / (-5+3)]
= 5/2
Slope of CA = [(4 + 3)/ (1 + 5)]
= 7/6
Thus, we have:
Slope of CF = -2
Slope of AD = -2/5
Slope of BE = -6/7
Hence,
Equation of CF is:
y + 3 = -2(x + 5)
y + 3 = -2x – 10
2x + y + 13 = 0
Equation of AD is:
y – 4 = (-2/5) (x – 1)
5y – 20 = -2x + 2
2x + 5y – 22 = 0
Equation of BE is:
y – 2 = (-6/7) (x + 3)
7y – 14 = -6x – 18
6x + 7y + 4 = 0
∴ The requiredequations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.